Tuple with Same Product

class Solution {
public:
    int countUniqueProductPairs(vector<int>& elements) {
        int arraySize = elements.size();
        unordered_map<int, int> productFrequencyMap;
            
        int tupleCount = 0;
    
        // Compute frequency of each product
        for (int i = 0; i < arraySize; i++) {
            for (int j = i + 1; j < arraySize; j++) {
                int product = elements[i] * elements[j];
                productFrequencyMap[product]++;
            }
        }
    
        // Count tuples for equivalent product pairs
        for (auto [product, count] : productFrequencyMap) {
            if (count > 1) {
                int pairs = count * (count - 1) / 2;
                tupleCount += 8 * pairs;
            }
        }
    
        return tupleCount;
    }
};

The solution provided in C++ aims to count the number of unique tuples where the products of pairs of integers from an array are equal. The process consists of two main steps:

1. Create and populate a frequency map to count occurrences of products of all possible pairs in the given input array.

2. Calculate and sum up the unique tuples based on the frequencies of these products, ensuring that each tuple consists of four elements (two pairs yielding the same product).

Below is a breakdown of how the code achieves this:

• Define a function countUniqueProductPairs which takes a vector of integers as an input.
• Calculate and store the size of the input array for efficient access.
• Use an unordered map (productFrequencyMap) to track the frequency of the product of any two elements.
• Iterate through all possible pairs in the input array using a nested loop structure. For each pair, the product is calculated and its occurrence is recorded in productFrequencyMap.
• After mapping, iterate through the productFrequencyMap to count tuples. If the frequency of a product is more than one (meaning at least two pairs have the same product), calculate the number of unique pairs & subsequently, the number of tuples using combinatorial mathematics.
• Finally, return the total count of such unique tuples.

This effective use of a hash map for product frequency serves to significantly optimize the counting process, allowing direct access to the number of times any particular product occurs, thereby facilitating efficient tuple counting.

class Solution {
    
    public int countTuplesWithSameProduct(int[] numsArray) {
        int numLength = numsArray.length;
    
        // Map to count each product's occurrence from pairs
        Map<Integer, Integer> productCountMap = new HashMap<>();
    
        int resultCount = 0;
    
        // Loop through the array to find and count products of pairs
        for (int i = 0; i < numLength; i++) {
            for (int j = i + 1; j < numLength; j++) {
                // Calculate product of two distinct elements
                int product = numsArray[i] * numsArray[j];
                productCountMap.put(product, productCountMap.getOrDefault(product, 0) + 1);
            }
        }
    
        // Calculate result based on counted products
        for (int product : productCountMap.keySet()) {
            int freq = productCountMap.get(product);
            // Compute combinations of pairs having the same product
            int pairsWithProduct = ((freq - 1) * freq) / 2;
            // Each valid pair combination corresponds to 8 tuples (as per problem statement)
            resultCount += 8 * pairsWithProduct;
        }
    
        return resultCount;
    }
}

In the Java solution for the "Tuple with Same Product" problem, the goal is to count tuples (groups of four numbers) from an array whose product pairs are the same. The implementation involves the use of a HashMap to record the frequency of each product formed by pairs of numbers from the array.

Here's how the algorithm works:

1. Define numsArray for storing the array of integers and calculate its length, numLength.
2. Create a HashMap, productCountMap, to keep track of how often each product appears when two different elements from the array are multiplied.
3. Initialize a counter, resultCount, to zero. This will store the final count of tuples.
4. Use a nested loop structure where the outer loop iterates through each element, and the inner loop pairs the current element from the outer loop with all subsequent elements.
5. For each pair, calculate the product and update productCountMap by either inserting the product as a new key with a value of 1 (if the product is not already in the map) or incrementing the existing value.
6. After all pairs have been processed, iterate over productCountMap to compute the count of tuples for each unique product. For each product, calculate the number of ways to pair elements (using combination formula) that have that product and then multiply this number by 8 (as each pair combination results in eight distinct tuples).
7. Add these calculated values to resultCount.
8. Finally, return resultCount as the output which is the total count of the tuples with the same product.

• Edge cases handled: The code effectively handles cases where products might clash by using a map to handle the frequency of each product.
• Efficiency: The solution is efficient in terms of space and time complexity due to the use of hashing for quick look-up and update operations.

This technique ensures that all possible pairs are considered without redundantly recalculating products, making the solution optimal for large datasets.

class Solution:
    def findTupleProductSum(self, elements):
        size = len(elements)
    
        # Dictionary to hold frequency of each product pair
        product_count = {}
    
        total_tups = 0
    
        # Check all pairs of 'elements'
        for i in range(size):
            for j in range(i + 1, size):
                # Compute the product of the pair
                product = elements[i] * elements[j]
                if product in product_count:
                    product_count[product] += 1
                else:
                    product_count[product] = 1
    
        # Calculate total based on counted frequencies
        for frequency in product_count.values():
            # Possible number of same product pair combinations
            combinations = (frequency - 1) * frequency // 2
    
            # Every combination can form 8 different tuples
            total_tups += 8 * combinations
    
        return total_tups

The solution involves calculating the number of unique tuples (a, b, c, d) such that the pairs (a, b) and (c, d) have the same product a * b = c * d, where the elements are distinct but the indices aren't necessarily. The algorithm takes the following steps using Python:

1. Define the function findTupleProductSum within a Solution class that accepts an array of integers named elements.

2. Initialize a dictionary product_count to track the frequency of each product computed from pairs of elements.

3. Initialize a variable total_tups to store the overall count of qualifying tuples.

4. Iterate over all possible pairs in the elements array using two nested loops. For each pair, compute the product. If this product already exists in product_count, increment its count; otherwise, add it to the dictionary with a count of 1.

5. After populating the dictionary, iterate over its values. For each product frequency, calculate the number of possible combinations using the combinations formula, which is derived as (frequency - 1) * frequency // 2. Multiply this number by 8 since each combination can create eight tuples based on the problem's symmetry conditions.

6. Return the accumulated count total_tups which represents the total number of qualifying tuples that can be formed from the elements array.

By ensuring each product of pairs is counted and subsequently calculating the number of tuples based on these counts, the program efficiently derives the solution to the problem. This approach minimizes unnecessary computations and optimally handles large input sizes.